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Unformatted text preview: Stat 219  Stochastic Processes Homework Set 8, Fall 2009, Due: November 18th 1. Exercise 5.2.5. Show that E ( , ) = by applying Doobs optional stopping theorem for the uniformly integrable stopped martingale W 2 t , t , . ANS: We have seen enroute to (5.2.2) that , < almost surely. Considering the martingale X t = W 2 t t of continuous sample path we have further assumed in the statement of the exercise that X t , is U.I. Thus, Doobs optional stopping theorem (Theorem 4.3.17) applies here, leading to the identity E ( W 2 , , ) = E ( W 2 0) = 0. That is, E , = E W 2 , = 2 P ( W , = ) + 2 P ( W , = ) = 2 + + 2 + = . 2. Exercise 5.2.7 (a) Suppose a random vector Y is independent of random vectors Z 1 ,Z 2 and Z 1 has the same law as Z 2 . Show that then Y + Z 1 has the same law as Y + Z 2 . ANS: A simple computation using characteristic functions and independence yields, E exp( i ( ,Y + Z 1 )) = E exp( i ( ,Y ) + i ( ,Z 1 )) = E exp( i ( ,Y )) E exp( i ( ,Z 1 )) (independence) = E exp( i ( ,Y )) E exp( i ( ,Z 2 )) ( Z 1 L = Z 2 ) = E exp( i ( ,Y ) + i ( ,Z 2 )) (independence) = E exp( i ( ,Y + Z 2 )) . We conclude that the characteristic functions of Y + Z 1 and Y + Z 2 agree and hence the two random vectors have the same distribution. (b) Suppose { Y t } and { Z t } are independent stochastic processes and { Z t } has the same distribution as { Z t } . Explain why X t = Y t Z t has the same distribution as W t = Y t + Z t . ANS: This immediately follows by applying part (a) to the finite dimensional distributions of { Y t } , { Z t } and { Z t } . (c) Let W t be a Brownian motion and 0 nonrandom. Upon verifying that part (b) applies for Y t = W t and Z t = W t W t deduce that X t = W t  ( W t W t ) is a Brownian motion. ANS: To apply (b) we need to show that { Y t } and { Z t } are independent stochastic processes. Let {G t } denote the canonical filtration of { W t } and note that is (trivially) a stopping time for {G t } . Note that Y t is measurable with respect to G for all t 0. Furthermore, Z t = 0 when t and Z t = W t W for t . Hence it follows from Proposition 5.2.3 that Z t is independent of {G } 1 and hence of Y t . Therefore by part (b), { X t } = { Y t Z t } L = { Y t + Z t } = { W t } , so that { X t } has the same finite dimensional distributions as a Brownian motion. Since its samplehas the same finite dimensional distributions as a Brownian motion....
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 Spring '09

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